Homework 1 Answer sheet Page 1 The SAS program I used to obtan the analyses for my answers s gven below. dm'log;clear;output;clear'; *************************************************************; *** EXST7034 Homework Example 1 ***; *** Problem from Neter, Wasserman & Kuttner 1989, #.18 ***; *************************************************************; OPTIONS LS=13 PS=56 NOCENTER NODATE NONUMBER nolabel; flename coper 'C:\Geaghan\Current\EXST7034\Fall005\SAS\CH01PR0.txt'; ODS HTML style=mnmal rs=none body='c:\geaghan\current\exst7034\fall005\sas\ch01pr0a.html' ; Ttle1 'Assgnment 1 : Coper mantanance example'; DATA ONE; INFILE coper MISSOVER; LABEL machnes = 'Number of machnes servced'; LABEL mnutes = 'Mnutes to servce machnes'; INPUT mnutes machnes; CARDS; RUN; ; optons ps=45; PROC PLOT DATA=ONE; PLOT mnutes * machnes; run; optons ps=56; PROC REG DATA=ONE lneprnter; ID machnes; MODEL mnutes = machnes / XPX I P; run; optons ps=55; plot predcted.*machnes='x' mnutes*machnes='o' / overlay; output out=next1 p=yhat r=e; run; optons ps=56; proc sort data=next1; by machnes; run; proc prnt data=next1; run; PROC REG DATA=ONE; MODEL mnutes = machnes; restrct ntercept = 0; run; PROC GLM; MODEL mnutes = machnes / XPX I P; run; A1 : Queston 1.1 n KNNL) The example gven was for sales dollar volume on the number of unts sold. If there was no source or errors, ths would be a functonal relatonshp, such that Y = X If, however, there were clercal errors n sales, the relatonshp would no longer be perfectly ftted by ths relatonshp. Although we may feel we know the underlyng functonal relatonshp, there would be uncertanty n the ft of the relatonshp. We would then ft the relatonshp usng Y = β 0 + β 1 X + ε for = 1,,, n where, Y = Dollar value of the sale X = Number of unts sold The random error term ε would address the uncertanty, and gve the varaton due to clercal errors. Addtonally, we mght expect β 0 to not dffer sgnfcantly from 0, gvng the relatonshp Y = β 1 X + ε. We may also hypothesze that β 1 does not dffer sgnfcantly from, f we feel that there s no bas or consstent tendency n the so called clercal errors, then the relatonshp s Y = X + ε. A : Queston 1. n KNNL) Ths functon would be would be fxed at Y = 300 + X, and would be a functonal relatonshp barrng any clercal errors. B1 : Queston 1.0a n KNNL) Obtan estmated regresson functon Y = 0.58016 + 15.0355X. Parameter Estmates Parameter Standard Varable DF Estmate Error t Value Pr > t Intercept 1-0.58016.80394-0.1 0.8371 machnes 1 15.0355 0.48309 31.1 <.0001
Homework 1 Answer sheet Page B1 : Queston 1.0b n KNNL) plot reg functon and data together Looks pretty good to me. Notce X has been used for the predcted values and o for the observed. The questons marks denote where SAS had to place BOTH a X and an o. --------+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-------- PRED 160 + + X 140 + + P? o r o o e d 10 + X + c t e d? 100 + o o + V a o o l o X u e 80 + +? o o o f m 60 +? + n u t X e 40 + o + s o o o? 0 + o + X 0 + + --------+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-------- 1.0 1.5.0.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 machnes B1 : Queston 1.0c n KNNL) gve an nterpretaton for b 0 The regresson functon s Y = 0.58016 + 15.0355X. The ntercept s theoretcal amount of tme requred to servce a machne when no machne s servced. We do not know how ths company blls for servce tme. Ths value could nclude travel tme, tme needed for settng up equpment before actually workng on a machne, tme to do paperwork after workng on a machne. If any of these are ncluded n the tme for a call, then we would expect the ntercept to be greater than zero, and t would estmated the tme needed (n mnutes) for these addton tasks. If on the other hand the servce tme ncludes only tme spent workng on a machne then we would expect the ntercept to be zero (e.g. no machne servced requres no tme). What we actually observe s a negatve number. If the real value s zero then we expect to see a small postve number about half the tme and a small negatve number about half the tme. The queston then becomes, does ths number dffer sgnfcantly from zero? SAS tests ths (see table for queston 1.0a above) and shows that the observed value does not dffer sgnfcantly from zero (P>F= 0.8371). If the hypotheses was rejected (H o : β 0 = 0) and the value was negatve then we may want to queston the model adequacy. The bottom lne: YES, I would say that the ntercept does tell us somethng about how ths company works and counts the tme recorded as servce tme. It may also tell us somethng about the best model (whch probably should go through the orgn).
Homework 1 Answer sheet Page 3 Note on testng parameters: Another way to test H o : β 0 = 0 s to ft a smple lnear regresson and restrct the model n SAS so that t forces the ntercept to be zero (e.g. PROC REG; MODEL Y=X; restrct ntercept=0;). SAS wll then report the parameter estmates (see output below) for the model wthout the ntercept (Y = 14.9473X.) and test the restrcton (P> t =0.8371). Parameter Estmates Parameter Standard Varable DF Estmate Error t Value Pr > t Intercept 1 1.0139E-14 0 Infty <.0001 machnes 1 14.9473 0.64 66.01 <.0001 RESTRICT -1-5.8680 8.0544-0.1 0.8371* * Probablty computed usng beta dstrbuton. B1 : Queston 1.0d n KNNL) Estmate servce tme when X = 5 Assgnment 1 : Coper mantanance example Obs machnes mnutes yhat e 1 1 1 14.455 -.4551 1 4 14.455-10.4551 3 1 3 14.455-11.4551 4 1 7 14.455 1.5449 5 0 9.490-9.4903 6 41 9.490 11.5097 7 3 9.490.5097 8 18 9.490-11.4903 9 0 9.490-9.4903 10 8 9.490-1.4903 11 34 9.490 4.5097 1 7 9.490 -.4903 13 3 46 44.56 1.4744 14 3 36 44.56-8.556 15 4 60 59.561 0.439 16 4 7 59.561 1.439 17 4 57 59.561 -.5608 18 4 57 59.561 -.5608 19 4 61 59.561 1.439 0 5 68 74.596-6.5961 1 5 89 74.596 14.4039 5 66 74.596-8.5961 3 5 74 74.596-0.5961 4 5 73 74.596-1.5961 5 5 90 74.596 15.4039 6 5 86 74.596 11.4039 7 5 77 74.596.4039 8 6 93 89.631 3.3687 9 6 96 89.631 6.3687 30 7 105 104.667 0.3334 31 7 101 104.667-3.6666 3 7 109 104.667 4.3334 33 7 11 104.667 7.3334 34 7 111 104.667 6.3334 35 7 11 104.667 7.3334 36 8 100 119.70-19.7018 37 8 131 119.70 11.98 38 8 13 119.70 3.98 39 9 144 134.737 9.69 40 9 134 134.737-0.7371 41 9 13 134.737 -.7371 4 9 131 134.737-3.7371 43 10 137 149.77-1.773 44 10 156 149.77 6.77 45 10 17 149.77 -.773 The values to the rght are output from PROC REG. PROC REG does not usually nclude the value of the ndependent varable n the output. The values of X have been ncluded n the output statstcs on the left because the varable X was ncluded n an ID statement. From the output statstcs t s clear that there were 8 observatons wth an X value equal to 5. For these values the pont estmate of servce tme, or predcted value, s equal to 74.596 mnutes.
Homework 1 Answer sheet Page 4 B : Queston 1.0c+ n KNNL addtonal request) gve an nterpretaton for b 1 In queston 1.0c we had a rather long-wnded nterpretaton of b 0. The value estmated for b 1 has a smpler nterpretaton. It s the change n servce tme per machne. Ths would be our best estmate of the tme requred to servce one machne, and was estmated as 15.0355 mnutes. B3 : ANOVA table) Ths was estmated by the SAS program as Analyss of Varance Sum of Mean Source DF Squares Square F Value Pr > F Model 1 76960 76960 968.66 <.0001 Error 43 3416.3770 79.45063 Corrected Total 44 80377 B4 : Normal Equatons) The normal equatons are provded n SAS n a somewhat ndrect form, but they can be obtaned. The matrx notaton for the normal equatons are (X'X)b = X'Y. The model opton XPX (e.g. MODEL Y=X / XPX;) wll cause the followng lstng. Model Crossproducts X'X X'Y Y'Y Varable Intercept machnes mnutes Intercept 45 30 343 machnes 30 1516 660 mnutes 343 660 3414 The 4 values n the upper-left porton of the lstng (18, 81, 81 and 439) are the elements of the X'X matrx and the two upper elements n the rght column (115 and 68) are the two values of the X'Y vector. The lower-rght element s the value of Y'Y. The reman two values (the frst two elements of the last row, 115 and 68) are the (X'Y) '. Ths causes the 9 values to create a symmetrc matrx. Multplyng out the matrx algebra brngs the equatons more n lne wth the algebrac verson of the normal equatons. The normal equatons expressed algebracally are: nb 0 + b 1 ΣX = ΣY b 0 ΣX + b 1 ΣX = ΣY X Fllng n the quanttatve values for the ntermedate calculatons we get: 45b 0 + 30b 1 = 343 30b 0 + 156b 1 = 660 These are the normal equatons, the equatons that must be solved to get estmates of b 0 and b 1. B5 : Regresson coeffcents and ther standard errors) Gven drectly from the SAS output Parameter Estmates Parameter Standard Varable DF Estmate Error t Value Pr > t Intercept 1-0.58016.80394-0.1 0.8371 machnes 1 15.0355 0.48309 31.1 <.0001
Homework 1 Answer sheet Page 5 B6 : Queston 1.4a n KNNL) obtan resduals and sum of square of resduals The resduals can be lsted n most SAS procedures (GLM, REG, MIXED, etc.). The resduals were lsted n queston 1.0d above. The sum of squared resduals are smply the SSError, and ths s an acceptable answer. However, PROC GLM actually calculates the sum of the resduals (squared and unsquared) f the P opton s specfed on the model. These values were gven as: Sum of Resduals 0.000000 Sum of Squared Resduals 3416.37703 Sum of Squared Resduals - Error SS 0.000000 The value mnmzed n fttng a least squares regresson s the sum of squares devatons (error). The book descrbes ths value as ( ) n Q= Y β β X = 1 n resduals are defned as ( ˆ ) equvalent. n = 1 = 1 0 1. Snce the sum of square of the devatons or e = Y Y, and ˆ Y = β0 β1x, the two values ( Q and e ) are Both were numercally equal to zero out to 6 decmal places (above). B6 : Queston 1.4b n KNNL) obtan estmates of σ and σ. These are estmated by the MSE from the SAS Analyss of Varance table (MSE = 79.45063) and the Root MSE (= 8.91351) also provded by SAS, usually followng the ANOVA table. The unts on the n = 1 Root MSE 8.91351 R-Square 0.9575 Dependent Mean 76.6667 Adj R-Sq 0.9565 Coeff Var 11.6879 estmate of σ would be the same as the dependent varable, mnutes. Addtonal questons for chapter were answer wth the followng statements. OPTIONS LS=111 PS=56 NOCENTER; PROC REG DATA=ONE lneprnter; ID machnes; MODEL mnutes = machnes / CLM CLI CLB alpha=0.10; output out=next1 p=yhat r=e; TEST machnes = 14; run; PROC GLM DATA=ONE; classes anotherx; MODEL mnutes = X anotherx; run; PROC MEANS DATA=ONE N MEAN SUM VAR USS CSS; VAR machnes mnutes; run; QUIT;
Homework 1 Answer sheet Page 6 Problem.5a) From the SAS output SAS provdes a confdence nterval for the regresson coeffcents (b 0 and b 1 ) and the value for α can be specfed, 0.10 n ths case. The resultng values were 14.314 and 15.84735. Jont confdence ntervals were not dscussed n chapter so I assume Parameter Estmates Parameter Standard Varable DF Estmate Error t Value Pr > t 90% Confdence Lmts Intercept 1-0.58016.80394-0.1 0.8371-5.9378 4.13347 machnes 1 15.0355 0.48309 31.1 <.0001 14.314 15.84735 they were not the ntended answer here. A more complete probablty statement would be gven as: P( b 1 -t α/ S b1 β 1 b 1 -t α/ S b1 ) = 1-α Where, n = 45 (d.f. = 43) and α = 0.10 so t α/ = 1.681070704 (from excel). From prevous work b 1 = 15.0355 and S b1 = 0.48309. P(15.0355 1.746*0.48309 β 1 15.0355 + 1.746*0.48309) = 0.90, and P(14.314155 β 1 15.84735845) = 0.90 Problem.5b) The t-test of a lnear assocaton s avalable n the SAS output (H 0 : β 1 =0). The alternatve s H 1 : β 1 0. Rejecton would result at the α = 0.10 level f the calculated value of the t- value was greater than the crtcal tabular value for 16 d.f., t = 1.681070704. SAS reports the actual calculated value to be 31.1 (from the table above) so we clearly reject the null hypothess. In the output above SAS also provdes a P value that ndcates that ths t-test s hghly sgnfcant (P> t )<0.0001. All ths means s that there appears to be a correlaton between these two varables, and that values of X would provde some utlty for estmatng Y. However, ths by no means ndcates that ths partcular lnear assocaton that we ftted (whch s lnear) s the best lnear model for the relatonshp or that a nonlnear model mght not be even better. Problem.5c) Snce the confdence nterval n part.5a does not nclude zero, and the results of the t-test n part.5b reject zero, the results are consstent. Problem.5d) Ths was done usng the TEST statement n SAS (.e. TEST X = 14;) However, ths s a two taled test. Snce we want a one taled test (α=0.05) we should look at the upper tal of only. Snce one tal equal to α=0.05 corresponds to a two taled test of α=0.10. Therefore, we would The REG Procedure Model: MODEL1 Test 1 Results for Dependent Varable mnutes Mean Source DF Square F Value Pr > F Numerator 1 364.8674 4.59 0.0378 Denomnator 43 79.45063 reject H 0 : β 1 =14 wth α=0.05 when the two taled test was less than P(>F)<0.10, AND IF THE RESULT WAS IN THE UPPER TAIL (.e. H 1 : β 1 > 14). To get a one taled P-value, dvde the observed P value by (e.g. P value = 0.0378 / = 0.0189). In ths case the P-value ndcates that the results are clearly n the upper tal. We would therefore reject the null hypothess (H 0 : β 1 =14 or H 0 : β 1 14), concludng that the alternatve hypothess (H 0 : β 1 >14) s a more reasonable concluson than the null hypothess. Ths s consstent wth the 90% confdence nterval for our estmate of β 1, P(14.314155 β 1 15.84735845) = 0.90.
Homework 1 Answer sheet Page 7 Problem.5e) Of course t does (dependng on what you call relevant nformaton). We EXPECT a value greater than zero f there s any startup tme for the machne repar. However, evdence ndcates that there s no start up tme. In the event there s no startup tme, then we mght expect the ntercept to not be sgnfcantly dfferent from zero (whch s the case for ths data). In the event that we should fnd a statstcally sgnfcantly negatve ntercept t mght suggest that the model s wrong or the data s wrong, snce for ths problem the ntercept should not be negatve. Problem.14a) The nformaton requested here s a 90% CLM. The SAS GLM output below contans ths confdence nterval as observaton 13 and observaton 30. Note that the value of X s lsted wth each observaton. The SAS statements producng ths output were: MODEL Y = X / CLM CLI CLB alpha=0.10; run; Queston.14a s answered by the results for observaton, where 6 machnes were servced. The answer to the queston s provded by the CLM (snce they ask about the mean servce tme) and the probablty statement s: P(87.839 E(Y X=6 ) 91.9788) = 0.90 Output Statstcs Dependent Predcted Std Error Obs machnes Varable Value Mean Predct 90% CL Mean 90% CL Predct Resdual 1 0.0000 9.4903.0061 6.1180 3.867 14.1313 44.8494-9.4903 4 60.0000 59.5608 1.4331 57.1517 61.9699 44.384 74.7375 0.439 3 3 46.0000 44.556 1.6750 41.7098 47.3414 9.791 59.771 1.4744 4 41.0000 9.4903.0061 6.1180 3.867 14.1313 44.8494 11.5097 5 1 1.0000 14.4551.3895 10.4381 18.471-1.058 9.9684 -.4551 6 10 137.0000 149.773.7099 145.168 154.378 134.1109 165.4337-1.773 7 5 68.0000 74.5961 1.398 7.3605 76.8316 59.4460 89.746-6.5961 8 5 89.0000 74.5961 1.398 7.3605 76.8316 59.4460 89.746 14.4039 9 1 4.0000 14.4551.3895 10.4381 18.471-1.058 9.9684-10.4551 10 3.0000 9.4903.0061 6.1180 3.867 14.1313 44.8494.5097 11 9 144.0000 134.7371.3011 130.8688 138.6054 119.616 150.16 9.69 1 10 156.0000 149.773.7099 145.168 154.378 134.1109 165.4337 6.77 13 6 93.0000 89.6313 1.3964 87.839 91.9788 74.4643 104.7983 3.3687 14 3 36.0000 44.556 1.6750 41.7098 47.3414 9.791 59.771-8.556 15 4 7.0000 59.5608 1.4331 57.1517 61.9699 44.384 74.7375 1.439 16 8 100.0000 119.7018 1.970 116.464 1.941 104.3714 135.03-19.7018 17 7 105.0000 104.6666 1.6119 101.9569 107.3763 89.4393 119.8939 0.3334 18 8 131.0000 119.7018 1.970 116.464 1.941 104.3714 135.03 11.98 19 10 17.0000 149.773.7099 145.168 154.378 134.1109 165.4337 -.773 0 4 57.0000 59.5608 1.4331 57.1517 61.9699 44.384 74.7375 -.5608 1 5 66.0000 74.5961 1.398 7.3605 76.8316 59.4460 89.746-8.5961 7 101.0000 104.6666 1.6119 101.9569 107.3763 89.4393 119.8939-3.6666 3 7 109.0000 104.6666 1.6119 101.9569 107.3763 89.4393 119.8939 4.3334 4 5 74.0000 74.5961 1.398 7.3605 76.8316 59.4460 89.746-0.5961 5 9 134.0000 134.7371.3011 130.8688 138.6054 119.616 150.16-0.7371 6 7 11.0000 104.6666 1.6119 101.9569 107.3763 89.4393 119.8939 7.3334 7 18.0000 9.4903.0061 6.1180 3.867 14.1313 44.8494-11.4903 8 5 73.0000 74.5961 1.398 7.3605 76.8316 59.4460 89.746-1.5961 9 7 111.0000 104.6666 1.6119 101.9569 107.3763 89.4393 119.8939 6.3334 30 6 96.0000 89.6313 1.3964 87.839 91.9788 74.4643 104.7983 6.3687 31 8 13.0000 119.7018 1.970 116.464 1.941 104.3714 135.03 3.98 3 5 90.0000 74.5961 1.398 7.3605 76.8316 59.4460 89.746 15.4039 33 0.0000 9.4903.0061 6.1180 3.867 14.1313 44.8494-9.4903 34 8.0000 9.4903.0061 6.1180 3.867 14.1313 44.8494-1.4903 35 1 3.0000 14.4551.3895 10.4381 18.471-1.058 9.9684-11.4551 36 4 57.0000 59.5608 1.4331 57.1517 61.9699 44.384 74.7375 -.5608 37 5 86.0000 74.5961 1.398 7.3605 76.8316 59.4460 89.746 11.4039 38 9 13.0000 134.7371.3011 130.8688 138.6054 119.616 150.16 -.7371 39 7 11.0000 104.6666 1.6119 101.9569 107.3763 89.4393 119.8939 7.3334 40 1 7.0000 14.4551.3895 10.4381 18.471-1.058 9.9684 1.5449 41 9 131.0000 134.7371.3011 130.8688 138.6054 119.616 150.16-3.7371 4 34.0000 9.4903.0061 6.1180 3.867 14.1313 44.8494 4.5097 43 7.0000 9.4903.0061 6.1180 3.867 14.1313 44.8494 -.4903 44 4 61.0000 59.5608 1.4331 57.1517 61.9699 44.384 74.7375 1.439 45 5 77 0000 74 5961 1 398 7 3605 76 8316 59 4460 89 746 4039
Homework 1 Answer sheet Page 8 Problem.14b) The nformaton requested here s a 90% CLI, snce t s a sngle trp. Ths s also avalable n the table above as observaton. The SAS GLM output below contans ths confdence nterval as observaton. The probablty statement s : P(74.4643 E(Y X=6 ) 104.7983) = 0.90 Problem.14c) The answer to queston.14a was P(87.839 E(Y X=6 ) 91.9788) = 0.90. Ths s the estmated tme for the repar of 6 machnes. Ths calculaton s done as a confdence nterval on the estmated value, b 1 ± t α/ S b1 = 89.6313 ± 1.681070704*1.3964. For ths problem the standard error 1 ( X X) s MSE + n. Σ( X X ) From proc means PROC MEANS DATA=ONE N MEAN SUM VAR USS CSS; VAR machnes mnutes; run; We get the result: The MEANS Procedure Varable N Mean Sum Varance USS Corrected SS ------------------------------------------------------------------------------- machnes 45 5.1111111 30.0000000 7.7373737 1516.00 340.4444444 mnutes 45 76.666667 343.00 186.75 3414.00 80376.80 -------------------------------------------------------------------------------- 1 ( X X) MSE + n s then 79.45063[1/45 + (X -5.1111111) /340.4444444]. For X = 6 ths value Σ( X X ) s 1.396410845. Ths value s provded by SAS (see STD ERROR MEAN PREDICT n the output statstcs). For a new sample of sze m = 6 the calculaton s 1 1 ( X X) MSE + + m n Σ( X X ) 1 ( X X) MSE MSE + n +. In order to estmate the mean tme for 6 machnes we need modfy Σ( X X ) m the prevous calculaton of the nterval of the mean by addng MSE/m = 79.45063/6 = 13.4177167. The varance for 6 machnes s then 79.45063[1/45 + (X -5.1111111) /340.4444444] + 13.4177167 = 15.19173491. Then the standard error s 15.19173491 = 3.897657619. The nterval s then b 1 ± t α/ S b1 = 89.6313 ± 1.681070704*3.897657619, and P(83.07906196 E(Y X=6 ) 96.18353804) = 0.90. = Problem.14d) Ths nterval for the mean of 6 machnes should be wder the nterval for the regresson lne, P(87.839 E(Y X=6 ) 91.9788) = 0.90, but narrower tan the nterval for ndvdual ponts, P(74.4643 E(Y X=6 ) 104.7983) = 0.90.